In this guide, you will get 189. Rotate Array LeetCode Solution with the best time and space complexity. The solution to Rotate Array problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Given an integer array nums, rotate the array to the right by k steps, where k is non-negative.
Example 1:
Input: nums = [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]
Example 2:
Input: nums = [-1,-100,3,99], k = 2
Output: [3,99,-1,-100]
Explanation:
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]
Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.
Could you do it in-place with O(1) extra space?
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(1)
189. Rotate Array LeetCode Solution in C++
class Solution {
public:
void rotate(vector<int>& nums, int k) {
k %= nums.size();
reverse(nums, 0, nums.size() - 1);
reverse(nums, 0, k - 1);
reverse(nums, k, nums.size() - 1);
}
private:
void reverse(vector<int>& nums, int l, int r) {
while (l < r)
swap(nums[l++], nums[r--]);
}
}; /* code provided by PROGIEZ */
189. Rotate Array LeetCode Solution in Java
class Solution {
public void rotate(int[] nums, int k) {
k %= nums.length;
reverse(nums, 0, nums.length - 1);
reverse(nums, 0, k - 1);
reverse(nums, k, nums.length - 1);
}
private void reverse(int[] nums, int l, int r) {
while (l < r)
swap(nums, l++, r--);
}
private void swap(int[] nums, int l, int r) {
final int temp = nums[l];
nums[l] = nums[r];
nums[r] = temp;
}
} // code provided by PROGIEZ
189. Rotate Array LeetCode Solution in Python
class Solution:
def rotate(self, nums: list[int], k: int) -> None:
k %= len(nums)
self.reverse(nums, 0, len(nums) - 1)
self.reverse(nums, 0, k - 1)
self.reverse(nums, k, len(nums) - 1)
def reverse(self, nums: list[int], l: int, r: int) -> None:
while l < r:
nums[l], nums[r] = nums[r], nums[l]
l += 1
r -= 1 # code by PROGIEZ
